“数组”: https://leetcode.com/tag/array/ “双指针”: https://leetcode.com/tag/two-pointers/ Similar Questions: “两数之和”: https://leetcode.com/problems/two-sum/ “最接近的三数之和”: https://leetcode.com/problems/3sum-closest/ “四数之和”: https://leetcode.com/problems/4sum/ “较小的三数之和”: https://leetcode.com/problems/3sum-smaller/

Problem: Link to heading

给你一个包含 n 个整数的数组 nums，判断 nums 中是否存在三个元素 *a，b，c ，*使得 *a + b + c =*0 ？请你找出所有和为 0 且不重复的三元组。

  • *注意：**答案中不可以包含重复的三元组。

示例 1：

输入：nums = [-1,0,1,2,-1,-4]
输出：[[-1,-1,2],[-1,0,1]]

示例 2：

输入：nums = []
输出：[]

示例 3：

输入：nums = [0]
输出：[]

提示：

  • 0 <= nums.length <= 3000
  • 105 <= nums[i] <= 105

Solution: Link to heading

https://leetcode-cn.com/problems/3sum/solution/san-shu-zhi-he-by-leetcode-solution/

class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        int n = nums.size();
        sort(nums.begin(), nums.end());
        vector<vector<int>> ans;
        // 枚举 a
        for (int first = 0; first < n; ++first) {
            // 需要和上一次枚举的数不相同
            if (first > 0 && nums[first] == nums[first - 1]) {
                continue;
            }
            // c 对应的指针初始指向数组的最右端
            int third = n - 1;
            int target = -nums[first];
            // 枚举 b
            for (int second = first + 1; second < n; ++second) {
                // 需要和上一次枚举的数不相同
                if (second > first + 1 && nums[second] == nums[second - 1]) {
                    continue;
                }
                // 需要保证 b 的指针在 c 的指针的左侧
                while (second < third && nums[second] + nums[third] > target) {
                    --third;
                }
                // 如果指针重合，随着 b 后续的增加
                // 就不会有满足 a+b+c=0 并且 b<c 的 c 了，可以退出循环
                if (second == third) {
                    break;
                }
                if (nums[second] + nums[third] == target) {
                    ans.push_back({nums[first], nums[second], nums[third]});
                }
            }
        }
        return ans;
    }
};